T-LP: Integration Tools for Sums, Products and Limits
This page supplies the analysis required for Chapter 15's maximal inequalities. The first three sections hold on arbitrary measure spaces; the last explicitly concerns probability spaces. Functions here are real-valued; completeness of the measure is not assumed, and functions are identified almost everywhere. These full written proofs remain subject to mathematical revision.
Why two different moments control a product
Let 1<p<∞ and q=p/(p−1). For a,b≥0, maximizing ab−a^p/p gives the maximizer a=b^{q-1}, hence
ab\le a^p/p+b^q/q.If f∈L^p and g∈L^q have nonzero norms, substitute a=|f|/‖f‖_p and b=|g|/‖g‖_q and integrate. The two terms sum to 1, giving
\int|fg|\,d\mu\le\|f\|_p\|g\|_q.If either norm is zero, its function vanishes almost everywhere and the conclusion still holds. For p=1 and q=∞, integrate the almost everywhere bound |g|≤‖g‖_∞ directly; exchange f and g for the other endpoint. This proves the required Hölder inequality on arbitrary measure spaces.
From a product estimate to the triangle inequality
For 1<p<∞, convexity first gives |f+g|^p\le2^{p-1}(|f|^p+|g|^p), so f+g belongs to L^p whenever f and g do. This first ensures that the following integrals make sense: we must not divide by a potentially infinite norm before establishing integrability. Set S=|f+g|. Hölder gives
\int S^p\,d\mu\le\int(|f|+|g|)S^{p-1}\,d\mu
\le(\|f\|_p+\|g\|_p)\|S\|_p^{p-1}.If ‖S‖_p>0, divide by the finite quantity \|S\|_p^{p-1} to obtain \|f+g\|_p\le\|f\|_p+\|g\|_p; a zero norm is immediate. For p=1 use the pointwise triangle inequality, and for p=∞ use essential bounds. This proves Minkowski for all 1≤p≤∞.
How a Cauchy sequence actually finds a measurable limit
Let 1≤p<∞ and let f_n be Cauchy in L^p. Extract a subsequence f_{n_j} with consecutive differences of L^p norm at most 2^{-j}. Finite applications of Minkowski followed by monotone convergence for the nonnegative integral give
\left\|\sum_{j=1}^m|f_{n_{j+1}}-f_{n_j}|\right\|_p
\le\sum_{j=1}^m2^{-j}\le1,
\qquad
\int\left(\sum_{j=1}^\infty|f_{n_{j+1}}-f_{n_j}|\right)^p d\mu\le1.Thus the absolute difference series is finite almost everywhere. Define f as f_{n_1} plus the sum of those consecutive differences, setting it to zero on the measurable exceptional null set. This is measurable; the same estimate and Minkowski show that f∈L^p. Apply Minkowski to finite tail sums and then Fatou to obtain
\|f-f_{n_j}\|_p\le\sum_{k=j}^\infty2^{-k}\longrightarrow0.Because the original sequence is Cauchy, for any ε>0 first choose j with subsequence error below ε/2, then use the Cauchy property to control f_n-f_{n_j}. The whole sequence converges to f. This proves Riesz–Fischer completeness for 1≤p<∞, rather than merely convergence of an extracted subsequence.
For p=∞, again extract a subsequence whose consecutive L∞ differences are at most 2^{-j}. Outside the union of their exceptional null sets, also removing the set where the first term exceeds its finite essential bound, all differences obey a uniformly summable bound, so the subsequence converges uniformly to a bounded function. Set it to zero on the measurable exceptional set to obtain a measurable essentially bounded representative. The tail estimate gives L∞ convergence, and the original Cauchy property gives convergence of the whole sequence. Thus L^p is complete for every 1≤p≤∞.
Approximation, conditional averaging and norm identification on a probability space
Here μ=P. For f∈L^p, 1≤p<∞, truncation f^{(K)}=f\mathbf1_{\{|f|\le K\}} satisfies \|f-f^{(K)}\|_p\to0, since its pth tail integral tends to zero. Approximate each bounded real measurable function by step functions of mesh 2^{-m}, with pointwise error at most 2^{-m}. Total probability 1 makes their L^p errors at most 2^{-m} as well. Bounded simple functions are therefore dense in these L^p spaces.
Hölder also gives L^p⊂L¹. Conditional Jensen from Chapter 9 and the order property of conditional averages consequently give contraction for every 1≤p≤∞:
\|\mathbb E[f\mid\mathcal G]\|_p\le\|f\|_p.For finite p, integrate |\mathbb E[f\mid\mathcal G]|^p\le\mathbb E[|f|^p\mid\mathcal G]. For p=∞, condition the bounds -\|f\|_\infty\le f\le\|f\|_\infty. At p=1 the conditional Jensen step is just the conditional absolute-value inequality.
Let q be the conjugate exponent, with 1 and ∞ conjugate. On this finite measure space, the norm-duality formula needed here is
\|f\|_p=\sup_{\|g\|_q\le1}\left|\mathbb E[fg]\right|.Hölder gives the upper bound. For 1<p<∞ and nonzero f, choose g=\operatorname{sgn}(f)|f|^{p-1}/\|f\|_p^{p-1} to attain equality; for p=1 use g=sgn(f). For p=∞ with \|f\|_\infty>0, take 0<ε<\|f\|_\infty and the positive-probability set A_ε={|f|>\|f\|_\infty−ε}. The choice g=\operatorname{sgn}(f)\mathbf1_{A_\varepsilon}/P(A_\varepsilon) reads at least \|f\|_\infty−ε; let ε decrease to zero. The zero function is immediate. This section identifies the norm only; it does not assert that every continuous linear functional on L∞ is represented by an L¹ function.